I'm carrying out a study on reported deaths of vultures (Gyps fulvus) in the Apennine mountains, Italy. I've performed a chi-square test on a 2x2 contingency table relating gender (male, female) to age (adult, subadult). Data are as follows: adult males= 8; adult females= 13; subadult males=13; subadult females= 4. The chi-square is 5.5981 p=0.018, so the two variables (gender, age) are not independent. Is there a method to calculate which observed value/s is/are significantly different from expected value/s? I.e. is the rejection of H0 depending on which particular cell value/s?

  • $\begingroup$ Calculating expected values from marginal totals leaves only one degree of freedom. There's no sense in saying the association is due to the high no. adult females when you could just as well say it's due to the high no. subadult males, or the low no. subadult females, &c. $\endgroup$ Jun 8, 2016 at 0:10
  • $\begingroup$ Thank you. So it a non sense for a 2x2 table to calculate standardized residuals and, accordingly, look for residuals values >2!? $\endgroup$
    – Mario
    Jun 9, 2016 at 9:12
  • $\begingroup$ All four standardized residuals have equal sizes: two are negative, the other two positive. $\endgroup$
    – whuber
    Jul 5, 2016 at 14:18

1 Answer 1


At the left is your table and at the right is a generic $2\times 2$ table showing the counts $a$ through $d$, marginal subtotals, and total $n=a+b+c+d$.

$$\begin{array}{r|rr|r} &\text{male} & \text{female} & \text{total} \\ \hline \text{adult} & 8 & 13 & 21\\ \text{subadult} & 13 & 4 & 17 \\ \hline \text{total} & 21 & 17 & 38 \end{array} \quad\quad \begin{array}{r|cc|r} &\text{column 1} & \text{column 2} & \text{total}\\ \hline \text{row 1} & a & b & a+b\\ \text{row 2} & c & d & c+d \\ \hline \text{total} & a+c & b+d & n \end{array} $$

In the $\chi^2$ analysis of this table, the row totals $a+b$ and $c+d$ are fixed and so are the column totals $a+c$ and $b+d$ (and so, therefore, must the total count $n$ be fixed). We contemplate varying the cell counts $a, b, c, d$ randomly subject to these constraints.

In this model, the total count acts like a conserved quantity in physics, such as the total amount of water in a (non-leaky) pipeline: counts can move around and reside in different parts of the system (the cells), but they cannot just disappear or grow.

The fixed column totals could be viewed as dual pistons, as shown in the figure.

Figure of two pistons representing a+c in one and b+d in the other.

These pistons are transparent so you can see the water and mechanical linkages within them. On the left there is $a$ amount of water at the top and $c$ at the bottom (just as the left column in the table has $a$ at the top and $c$ at the bottom). These lumps of water are confined by solid disks connected by a tie rod of fixed length. (Its length is a property of the system; it could change from one problem to the next.) On the right there similarly is $b$ water at the top and $d$ at the bottom, separated by two disks and a fixed-length tie rod of its own.

If the pistons were sealed, nothing could change. But these are connected at their tops and bottoms by two narrow pipes. Those pipes permit water to flow between $a$ and $b$ and between $c$ and $d$. We may therefore vary, say, the amount $a$ by pushing or pulling on the left tie rod. Pushing it up, for example, causes water to flow out of $a$ and into $b$. However, because the total amount of water in any isolated part of the system is conserved, the total in $a$ and $b$ remains constant. In the example, that total would always be $21$, the first row sum.

Similarly, the bottom pipe permits flow between $c$ and $d$, but conserves the total $c+d=17$ (the second row sum).

In this fashion these two connected dual pistons physically enforce the four marginal constraints in the table: all row sums and all column sums remain constant.

The "one degree of freedom" is afforded by our ability to move either of the tie rods. When we do so, the water flowing out one pipe and back in through the other forces the tie rod in the other piston to move in the opposite direction.

Figure 2, showing the system after $a$ has been reduced.

In this new configuration, the left tie rod has been pushed upward, reducing $a$ and thereby increasing $b$ and $c$ at the expense of $d$.

Obviously, the changes in the amounts of water in the four regions are all of the same size--but negative amounts for $a$ and $d$ and positive amounts for $b$ and $c$.

The first figure shows the system at "equilibrium" where the relative amounts of water in each piston are the same: $a:c$ equals $b:d$. These are the "expected values" of the table. The amount of flow that occurred in going from the first figure to the second affects $a$, $b$, $c$, and $d$ equally. These changes are the residuals. The test of independence in a two-way table adopts the null hypothesis that the pistons have varied only randomly from their equilibrium position. The $\chi^2$ statistic measures how much variation there has been. Exactly the same amount of variation occurs in each cell (up to sign).

Our conclusion is that because water (=counts) is conserved, all residuals are equal in size. Furthermore, the residuals for $a$ and $d$ have the same sign, which is opposite the sign of the residuals for $b$ and $c$. Consequently, we cannot attribute any deviation from equilibrium to any particular cell: they have all been affected equally.


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